Bạn tham khảo:
Ta có $\dfrac{a}{b}=$$\dfrac{b}{c}=$$\dfrac{c}{a}$
⇒$a^{2}=bc;b^2=ac;c^2=ab$
$P=\dfrac{a^3+b^3+c^3}{3abc}$
$P=\dfrac{a^3}{3abc}+$$\dfrac{b^3}{3abc}+$$\dfrac{c^3}{3abc}$
$P=\dfrac{a^2}{3bc}+$$\dfrac{b^2}{3ac}+ $$\dfrac{c^2}{3ab}$
$P=\dfrac{a^2}{3a^2}+$$\dfrac{b^2}{3b^2}+$$\dfrac{c^2}{3c^2}$
$P=\dfrac{1}{3}+$$\dfrac{1}{3}+$$\dfrac{1}{3}=1$
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